Q110 / 155 · WBJEE 2022 · Mathematics

From a balloon rising vertically with uniform velocity $\mathrm{ft} / \mathrm{sec}$ a piece of stone is let go. The height of the balloon above the ground when the stone reaches the ground after $4 \mathrm{sec}$ is $\left[\mathrm{g}=32 \mathrm{ft} / \mathrm{sec}^2\right]$

  • A
    $220 \mathrm{ft}$
  • B
    $240 \mathrm{ft}$
  • Correct answer
    C
    $256 \mathrm{ft}$
  • D
    $260 \mathrm{ft}$

Solution

When stone is let go, its velocity $=-v$ (downwards) Let, it was at a height h. $\therefore h=-v t+\frac{1}{2} gt^2$ $\begin{aligned} &=-v \times 4+\frac{1}{2} \times 32 \times 4^2 \\ &=-4 v+256 \end{aligned}$ Balloon reaches $4 \mathrm{vtt}$ in 4 seconds $\therefore$ height required $=4 v+h=256 \mathrm{ft}$

Hence, the correct answer is option: 'C'.

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